4x+38=2x2

Solution for 4x+38=2x2 equation:

4x+38=2x^2

We move all terms to the left:

4x+38-(2x^2)=0

determiningTheFunctionDomain
-2x^2+4x+38=0

a = -2; b = 4; c = +38;
Δ = b2-4ac
Δ = 42-4·(-2)·38
Δ = 320
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8\sqrt{5}}{2*-2}=\frac{-4-8\sqrt{5}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8\sqrt{5}}{2*-2}=\frac{-4+8\sqrt{5}}{-4}
$